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void f(int* begin, int* end) { std::fill(begin, end, 0); }
which turns into a != exit expression where one pointer is
strided and (thanks to step #1) known to not overflow, and
the other is loop invariant.
The observation here is that, though the IV is strided by
4 in this case, that the IV *has* to become equal to the
end value. It cannot "miss" the end value by stepping over
it, because if it did, the strided IV expression would
eventually wrap around.
Handle this by turning A != B into "A-B != 0" where the A-B
part is known to be NUW.
llvm-svn: 123131
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